What is the sum from n=1 to infinity of (n^2)/(5n) ?

The sum from n=1 to infinity of (n^2)/(5n) is diverges

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sum from n=1 to infinity of (n^2)/(5n)

n = 1 \sum \infty \frac{n ^{2}}{5 n}

diverges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{5 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{n ^{2}}{n}

Apply Series Divergence Test:

diverges

= \frac{1}{5} diverges

= diverges

\frac{d}{d t} (\frac{1}{1 + t})

f (x) = (x + 2)^{\frac{1}{2}}

n \to \infty lim (\frac{( 5 n )}{5 n + 7})

\frac{\partial}{\partial y} (\frac{9 y}{x})

\int sin (k x) d x

What is the sum from n=1 to infinity of (n^2)/(5n) ?
The sum from n=1 to infinity of (n^2)/(5n) is diverges